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valentina_108 [34]
3 years ago
15

A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 t. What minimum current in the rod is nee

ded in order for the magnetic force to balance the weight of the rod?
Physics
1 answer:
slamgirl [31]3 years ago
8 0

Answer:

i = 4.9 A

Explanation:

Force on a current carrying rod due to magnetic field is given as

F = iLB

here we know that

i =current in the rod

B = 0.10 T

L = 1.0 m

now magnetic force is balanced by the weight of the rod

so we will have

iLB = mg

i(1.0)(0.10) = 0.05 \times 9.8

i = 4.9 A

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Which has greater buoyant force, light corn syrup or the vegetable oil?
natta225 [31]
It is most likely vegetable oil
6 0
4 years ago
An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N
algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

4 0
3 years ago
A motorbike reaches a speed of 20 m/s over 60m, whilst
Fynjy0 [20]

Initial speed = 2√10 m/s

<h3>Further explanation  </h3>

Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration  

An equation of uniformly accelerated motion  

V = vo + at  

Vt² = vo² + 2a (x-xo)  

x = distance on t  

vo / vi = initial speed  

vt / vf = speed on t / final speed  

a = acceleration  

vf=20 m/s

d = 60 m

a = 3 m/s²

\tt vf^2=vi^2+2.ad\\\\20^2=vi^2+2\times 3\times 60\\\\400=vi^2+360\\\\40=vi^2\\\\vi=\sqrt{40}=2\sqrt{10}~m/s

7 0
3 years ago
In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48
cluponka [151]

Answer:

The thickness is  \Delta y =  2.4 *10^{-6} \  m

Explanation:

From the question we are told that

   The wavelength is  \lambda  = 480 \ nm = 480*10^{-9} \  m

    The first order of the dark  fringe is  m_1 =  16

     The second order of dark fringe considered is  m_2 = 6

Generally the condition for destructive interference is mathematically represented as

        y = \frac{m \lambda}{2}

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So  the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

      y_1 - y_2 = \Delta y =  \frac{m_1 \lambda}{2} -  \frac{m_2 \lambda}{2}

=>  y_1 - y_2 = \Delta y =  \frac{16 *480*10^{-9}}{2} -  \frac{6 *480*10^{-9}}{2}

=>  y_1 - y_2 = \Delta y =  5 (480*10^{-9})

=>  \Delta y =  2.4 *10^{-6} \  m

8 0
3 years ago
Plants appear green because they do not absorb the green wavelengths of light what happens to those green light waves when they
Arlecino [84]
The green wavelengths are reflected, causing green to be the only color we see in certain parts of plants.

Best of luck, my man.
4 0
3 years ago
Read 2 more answers
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